QNEET-2013-Ph-1507: NEET 2013 Physics - Class 12 NEET-2013-Ph-1507 Solution: Two pith balls carrying equal charges are suspended from a…

1. Set up the Equilibrium Conditions for Case 1: Let the vertical distance from the suspension point to the line joining the balls be $y$, the string length be $l$, and the angle with the vertical be $\theta$. Three forces act on each pith ball: � Tension ($T$) along the string � Weight ($mg$) pointing straight down � Electrostatic Repulsive Force ($F$) acting horizontally: $F=\frac{q^{2}}{4\pi \varepsilon _{0}r^{2}}$ Resolving components at equilibrium: $T\cos \theta =mg$ $T\sin \theta =F$ Dividing the two equations gives: $\tan \theta =\frac{F}{mg}$ 2. Apply the Small-Angle Approximation: For small values of $\theta$, we can approximate $\tan\theta \approx \frac{\text{Opposite}}{\text{Adjacent}} = \frac{r/2}{y}$: $\frac{r}{2y}=\frac{q^{2}}{4\pi \varepsilon _{0}r^{2}mg}\implies y=\left(\frac{2\pi \varepsilon _{0}mg}{q^{2}}\right)r^{3}$ This establishes a clean mathematical proportionality: $y\propto r^{3}\quad \text{or}\quad r\propto y^{1/3}$ 3. Analyze Case 2 (Clamped at half the height): When the strings are fixed at half their original vertical suspension height, the new height becomes $y' = \frac{y}{2}$. Letting the new separation be $r^{\prime }$: $\frac{y^{\prime }}{y}=\left(\frac{r^{\prime }}{r}\right)^{3}$ Substitute $y' = \frac{y}{2}$: $\frac{y/2}{y}=\left(\frac{r^{\prime }}{r}\right)^{3}\implies \frac{1}{2}=\left(\frac{r^{\prime }}{r}\right)^{3}$ Taking the cube root of both sides gives the final value: $\frac{r^{\prime }}{r}=\frac{1}{\sqrt[3]{2}}\implies r^{\prime }=\frac{r}{\sqrt[3]{2}}$